Đặt \(N=\sqrt{4+\sqrt{10+2\sqrt{5}}}+\sqrt{4-\sqrt{10+2\sqrt{5}}}\)
\(\Rightarrow N^2=4+\sqrt{10+2\sqrt{5}}+4-\sqrt{10+2\sqrt{5}}\)(N > 0)\(+2\sqrt{\left(4+\sqrt{10+2\sqrt{5}}\right)\left(4-\sqrt{10+2\sqrt{5}}\right)}\)(mình viết k đủ nên xuống dòng)
\(\Leftrightarrow N^2=8+2\sqrt{4^2-\left(10+2\sqrt{5}\right)^2}\)
\(\Leftrightarrow N^2=8+2\sqrt{4^2-10-2\sqrt{5}}\)
\(\Leftrightarrow N^2=8+2\sqrt{6-2\sqrt{5}}\)
\(\Leftrightarrow N^2=8+2.\left(\sqrt{5}-1\right)\)
\(\Leftrightarrow N^2=8+2\sqrt{5}-2\)
\(\Leftrightarrow N^2=6+2\sqrt{5}\)
\(\Rightarrow N=\sqrt{6+2\sqrt{5}}\)
\(\Rightarrow N=\sqrt{5}+1\)
(Nhớ k cho mình với nhoa!)
đặt biểu thức trên là A ta có:
A^2=\(\left(\sqrt{4+\sqrt{10+2\sqrt{5}}+\sqrt{4-\sqrt{10+2\sqrt{5}}}}\right)^2\)
=\(4+\sqrt{10+2\sqrt{5}}+2\sqrt{4^2-\left(10-2\sqrt{5}\right)^2}+4-\sqrt{10-2\sqrt{5}}\)
=\(8+2\sqrt{6-2\sqrt{5}}\)
=\(8+2\left(\sqrt{5}-1\right)\)
=\(6+2\sqrt{5}\)
=\(\left(\sqrt{5}+1\right)^2\)
=> A=\(\sqrt{\left(\sqrt{5}+1\right)^2}\)
=> A=\(\sqrt{5}-1\)vì A lớn hơn 0
